Problem: Divide the following complex numbers. $ \dfrac{-15+12i}{-4-5i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-4+5i}$ $ \dfrac{-15+12i}{-4-5i} = \dfrac{-15+12i}{-4-5i} \cdot \dfrac{{-4+5i}}{{-4+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-15+12i) \cdot (-4+5i)} {(-4-5i) \cdot (-4+5i)} = \dfrac{(-15+12i) \cdot (-4+5i)} {(-4)^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-15+12i) \cdot (-4+5i)} {(-4)^2 - (-5i)^2} = $ $ \dfrac{(-15+12i) \cdot (-4+5i)} {16 + 25} = $ $ \dfrac{(-15+12i) \cdot (-4+5i)} {41} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-15+12i}) \cdot ({-4+5i})} {41} = $ $ \dfrac{{-15} \cdot {(-4)} + {12} \cdot {(-4) i} + {-15} \cdot {5 i} + {12} \cdot {5 i^2}} {41} $ Evaluate each product of two numbers. $ \dfrac{60 - 48i - 75i + 60 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{60 - 48i - 75i - 60} {41} = \dfrac{0 - 123i} {41} = -3i $